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Continuity And Differentiability

Question

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Differentiate the following functions w.r.t. x : $straight x to the power of straight x plus straight x to the power of log space straight x end exponent$

Solution

Let space space straight y equals straight x to the power of straight x plus straight x to the power of log space straight x end exponent Put space straight x to the power of straight x equals straight u comma space straight x to the power of log space straight x end exponent equals straight v therefore space straight y equals straight u plus straight v therefore space dy over dx equals du over dx plus dv over dx space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis Now space straight u equals straight x to the power of straight x therefore space log space straight u equals log space straight x to the power of straight x rightwards double arrow space log space straight u equals straight x. log space straight x therefore space 1 over straight u du over dx equals straight x.1 over straight x plus log space straight x.1 therefore space du over dx equals straight u left parenthesis 1 plus log space straight x right parenthesis therefore space du over dx equals straight x to the power of straight x left parenthesis 1 plus log space straight x right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis Also space straight v equals straight x to the power of log space straight x end exponent therefore space log space straight v equals log left parenthesis straight x to the power of log space straight x end exponent right parenthesis rightwards double arrow space log space straight v equals log space straight x. log space straight x therefore space 1 over straight v dv over dx equals log space straight x.1 over straight x plus log space straight x.1 over straight x therefore space dv over dx equals straight v open parentheses fraction numerator 2 log space straight x over denominator straight x end fraction close parentheses therefore space dv over dx equals 2 fraction numerator log space straight x over denominator straight x end fraction. straight x to the power of log space straight x end exponent therefore space dv over dx equals 2 log space straight x. straight x to the power of lox space minus 1 end exponent space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis

From (1), (2) and (3), we get,

space space space space space space dy over dx equals straight x to the power of straight x left parenthesis 1 plus log space straight x right parenthesis plus 2 log space straight x. straight x to the power of log space straight x minus 1 end exponent

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